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I want to know two things: 1.Given the equation of a plane curve ,how do we know whether it is closed or not ?To be specific ,consider the following $$(x+y)^{2}+(x-y)^{\left(\frac{2}{3}\right)}=1$$.I drew the above curve in desmos url{https://www.desmos.com/calculator/z39avcy2er}and found it be a nice closed curve. Second,how can I find the length and area of such a curve as the above example? Thanks for any hints/respences.

AgnostMystic
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3 Answers3

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I'm taking a slightly different approach to that of @Raffaele because I recognize the form as one for which we can obtain an analytic solution. We begin by simplifying the equation by rotating the axes by $\theta=-\pi/4$, thus

$$ x=x'\cos(\theta)-y'\sin(\theta)=\frac{\sqrt{2}}{2}(x'+y')\\ y=x'\sin(\theta)+y'\cos(\theta)=\frac{\sqrt{2}}{2}(-x'+y')\\ $$

I follows that

$$ x+y=\sqrt{2}y'\\ x-y=\sqrt{2}x'\\ (\sqrt{2}x')^{2/3}+(\sqrt{2}y')^2=1 $$

We recognize this to be in the form superconics that we developed (see, for example here). We can write the above equation more generally in the canonical form

$$ X^q+Y^{1/p}=1\\ \text{or}\\ f(X)=(1-|X|^q)^p $$

where $X=x'/a,\ Y=y'/b,\ a=b=1/\sqrt{2},\ q=2/3,\ p=1/2$ and $-1\le X\le 1$ and $Y\ge 0$.

The important point here is that we can solve for the area analytically. The area for the closed curve is shown as follows

$$ \begin{align} A&=2ab\int_{-1}^1 f(X)\ dX\\ &=2ab\int_{-1}^1 (1-|X|^q)^p\ dX\\ &=4ab\int_{0}^1 (1-X^q)^p\ dX\\ &=4ab\frac1qB(1/q,p+1,1)\\ &=4ab\frac{\Gamma(p+1)\Gamma(1+1/q)}{\Gamma(p+1+1/q)}\\ &=3\pi/8 \end{align} $$

where $B$ is the beta function and $\Gamma$ is the gamma function. The numerical result is for the parameters of the present problem.

No such good fortune for the arc length, however. Typically we resort to numerical methods. Thus we solved

$$ \begin{align} L&=4a\int_0^1 \sqrt{1+f'(X)^2}\ dX\\ &=2\sqrt{2}\int_0^1 \sqrt{1+\frac{1}{9(1-X^{2/3})X^{2/3}}}\ dX\\ &=4.1217... \end{align} $$

Cye Waldman
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Starting from @Cye Waldman's answer $$L=2\sqrt{2}\int_0^1 \sqrt{1+\frac{1}{9(1-x^{2/3})x^{2/3}}}\ dx$$ we can make an exact calculation.

Let $x=\sin^3(t)$ to get $$L=\int_0^\frac \pi 2 \sin(t)\sqrt{8+18 \sin^2(2 t)} \, dt$$ This leads to awful elliptic integrals of the first and second kind.

Making the story short, the result is

$$\frac{i \left(\frac{9 }{\sqrt{2}}L-6\right)}{\sqrt{6 \left(3+\sqrt{13}\right)}}=$$ $$\sqrt{13} F\left(i \sinh ^{-1}\left(\sqrt{\frac{3}{2} \left(3+\sqrt{13}\right)}\right)|\frac{-11+3 \sqrt{13}}{2} \right)-3 E\left(i \sinh ^{-1}\left(\sqrt{\frac{3}{2} \left(3+\sqrt{13}\right)}\right)|\frac{-11+3 \sqrt{13}}{2} \right)$$ which makes $$L=4.1216964106052626439040756172731180548685478808226\cdots$$ which is not recognized by inverse symbolic calculators.

Claude Leibovici
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  • Claude,this is very impressive. I wonder if there is a possibility of finding the arc length for a general superconics profile, say $f(x)=(1-|x|^q)^p$. These are symmetric about both axes, so we can give the derivative in the first quadrant as $f'(x)=-pqx^{q-1}(1-x^q)^{p-1}$. If this is too much to chew, how about the special cases of the superellipse ($p=1/q$) or superparabola ($q=2$). The area, centroid, and moments of these forms have already been determined in terms of gamma functions. If you wish, I can post this as a separate question. Cye – Cye Waldman Oct 19 '20 at 16:02
  • @CyeWaldman. Good idea to post the problem. Cheers :-) – Claude Leibovici Oct 19 '20 at 18:09
  • Real-only solution: $L=\sqrt{6\sqrt{13}}(2E(\psi,m)/3+(\sqrt{13}-3)F(\psi,m)/9)+D$ where $\sin^2\psi=(33\sqrt{13}-117)/2$, $m=1/2+3\sqrt{13}/26$ and $D=\sqrt{476-132\sqrt{13}}/3$. – Parcly Taxel Dec 11 '20 at 16:02
  • *Found with Byrd and Friedman 235.11 and 361.16. – Parcly Taxel Dec 11 '20 at 16:06
  • @ParclyTaxGood to know it ! Thanks for telling. Cheers :-) – Claude Leibovici Dec 11 '20 at 16:50
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The curve $$(x-y)^{2/3}+(x+y)^2=1$$ rotated of $\pi/4$ with the transformation $$\left\{x\to \frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}},y\to \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right\}$$ becomes $$2 x^2+\sqrt[3]{2} \,y^{2/3}=1$$ For the simmetry of the curve, the area is given by $$S=4 \int_0^{\frac{1}{\sqrt{2}}} \frac{\left(1-2 x^2\right)^{3/2}}{\sqrt{2}} \, dx=\frac{3 \pi }{8}$$ While the curve length leads to an elliptic integral $$L=4\int_0^{\frac{1}{\sqrt{2}}} \sqrt{-36 x^4+18 x^2+1} \, dx\approx 4.12$$

Edit.

I forgot to prove that this is a closed curve.

Solve the equation wrt $x$ $$x=\pm\frac{\sqrt{1-\sqrt[3]{2} y^{2/3}}}{\sqrt{2}}$$ in order to be real we must have $$\sqrt{1-\sqrt[3]{2} y^{2/3}}\ge 0 \to 0\leq y\leq \frac{1}{\sqrt{2}}$$ and so the limitations for $x$ are $$-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}$$

Raffaele
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